A sequence {x1, x2, .. xn} is alternating sequence if its elements satisfy one of the following relations :

x1 < x2 > x3 < x4 > x5 < …. xn or x1 > x2 < x3 > x4 < x5 > …. xn

Examples:

Input: arr[] = {1, 5, 4} Output: 3 The whole arrays is of the form x1 < x2 > x3 Input: arr[] = {1, 4, 5} Output: 2 All subsequences of length 2 are either of the form x1 < x2; or x1 > x2 Input: arr[] = {10, 22, 9, 33, 49, 50, 31, 60} Output: 6 The subsequences {10, 22, 9, 33, 31, 60} or {10, 22, 9, 49, 31, 60} or {10, 22, 9, 50, 31, 60} are longest subsequence of length 6.

This problem is an extension of longest increasing subsequence problem, but requires more thinking for finding optimal substructure property in this.

We will solve this problem by dynamic Programming method, Let A is given array of length n of integers. We define a 2D array las[n][2] such that las[i][0] contains longest alternating subsequence ending at index i and last element is greater than its previous element and las[i][1] contains longest alternating subsequence ending at index i and last element is smaller than its previous element, then we have following recurrence relation between them,

las[i][0]= Length of the longest alternating subsequence ending at index i and last element is greater than its previous elementlas[i][1]= Length of the longest alternating subsequence ending at index i and last element is smaller than its previous elementRecursive Formulation:las[i][0] = max (las[i][0], las[j][1] + 1); for all j < i and A[j] < A[i] las[i][1] = max (las[i][1], las[j][0] + 1); for all j < i and A[j] > A[i]

The first recurrence relation is based on the fact that, If we are at position i and this element has to bigger than its previous element then for this sequence (upto i) to be bigger we will try to choose an element j ( < i) such that A[j] < A[i] i.e. A[j] can become A[i]’s previous element and las[j][1] + 1 is bigger than las[i][0] then we will update las[i][0].

Remember we have chosen las[j][1] + 1 not las[j][0] + 1 to satisfy alternate property because in las[j][0] last element is bigger than its previous one and A[i] is greater than A[j] which will break the alternating property if we update. So above fact derives first recurrence relation, similar argument can be made for second recurrence relation also.

## C

// C program to find longest alternating subsequence in // an array #include <stdio.h> #include <stdlib.h> // function to return max of two numbers int max(int a, int b) { return (a > b) ? a : b; } // Function to return longest alternating subsequence length int zzis(int arr[], int n) { /*las[i][0] = Length of the longest alternating subsequence ending at index i and last element is greater than its previous element las[i][1] = Length of the longest alternating subsequence ending at index i and last element is smaller than its previous element */ int las[n][2]; /* Initialize all values from 1 */ for (int i = 0; i < n; i++) las[i][0] = las[i][1] = 1; int res = 1; // Initialize result /* Compute values in bottom up manner */ for (int i = 1; i < n; i++) { // Consider all elements as previous of arr[i] for (int j = 0; j < i; j++) { // If arr[i] is greater, then check with las[j][1] if (arr[j] < arr[i] && las[i][0] < las[j][1] + 1) las[i][0] = las[j][1] + 1; // If arr[i] is smaller, then check with las[j][0] if( arr[j] > arr[i] && las[i][1] < las[j][0] + 1) las[i][1] = las[j][0] + 1; } /* Pick maximum of both values at index i */ if (res < max(las[i][0], las[i][1])) res = max(las[i][0], las[i][1]); } return res; } /* Driver program */ int main() { int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 }; int n = sizeof(arr)/sizeof(arr[0]); printf("Length of Longest alternating subsequence is %d\n", zzis(arr, n) ); return 0; }

## Java

// Java program to find longest // alternating subsequence in an array import java.io.*; class GFG { // Function to return longest // alternating subsequence length static int zzis(int arr[], int n) { /*las[i][0] = Length of the longest alternating subsequence ending at index i and last element is greater than its previous element las[i][1] = Length of the longest alternating subsequence ending at index i and last element is smaller than its previous element */ int las[][] = new int[n][2]; /* Initialize all values from 1 */ for (int i = 0; i < n; i++) las[i][0] = las[i][1] = 1; int res = 1; // Initialize result /* Compute values in bottom up manner */ for (int i = 1; i < n; i++) { // Consider all elements as // previous of arr[i] for (int j = 0; j < i; j++) { // If arr[i] is greater, then // check with las[j][1] if (arr[j] < arr[i] && las[i][0] < las[j][1] + 1) las[i][0] = las[j][1] + 1; // If arr[i] is smaller, then // check with las[j][0] if( arr[j] > arr[i] && las[i][1] < las[j][0] + 1) las[i][1] = las[j][0] + 1; } /* Pick maximum of both values at index i */ if (res < Math.max(las[i][0], las[i][1])) res = Math.max(las[i][0], las[i][1]); } return res; } /* Driver program */ public static void main(String[] args) { int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 }; int n = arr.length; System.out.println("Length of Longest "+ "alternating subsequence is " + zzis(arr, n)); } } // This code is contributed by Prerna Saini

Output:

Length of Longest alternating subsequence is 6

Time Complexity: O(n^{2})

Auxiliary Space: O(n)

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.